The area of 3 adjacent faces of a cuboid are x,y,z.. if the volume is v , prove that [v]^2=xyz.

To prove that [v]^2 = xyz, where x, y, and z are the areas of three adjacent faces of a cuboid, and v is its volume, we will use the formula for the volume of a cuboid.

Let's consider a cuboid with length L, width W, and height H. The volume V of this cuboid is given by V = L * W * H.

Now, let's find the areas of the three adjacent faces. We can calculate these by multiplying the respective dimensions of the cuboid. Assume that x is the area of the face with dimensions L and W, y is the area of the face with dimensions W and H, and z is the area of the face with dimensions L and H.

Therefore, we have:
x = L * W
y = W * H
z = L * H

Now, let's substitute the values of x, y, and z into the equation [v]^2 = xyz and see if it holds true:

[v]^2 = (L * W * H) * (L * W) * (W * H) = L^2 * W^2 * H^2

Comparing this with xyz, we can see that [v]^2 = xyz.

Hence, we have proved that [v]^2 = xyz, where x, y, and z are the areas of three adjacent faces of a cuboid, and v is its volume.

if the edges are a,b,c

x = ab
y = bc
z = ac

v = abc, so

v^2 = (abc)^2
= a^2 b^2 c^2
= ab * bc * ac
= xyz

thank you